Solve by Factoring 11r+15=-2r^2

$11 r + 15 = - 2 r^{2}$

Step 1

Add $2 r^{2}$ to both sides of the equation.

$11 r + 15 + 2 r^{2} = 0$

Step 2

Factor by grouping.

Reorder terms.

$2 r^{2} + 11 r + 15 = 0$

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 15 = 30$ and whose sum is $b = 11$ .

Factor $11$ out of $11 r$ .

$2 r^{2} + 11 (r) + 15 = 0$

Rewrite $11$ as $5$ plus $6$

$2 r^{2} + (5 + 6) r + 15 = 0$

Apply the distributive property.

$2 r^{2} + 5 r + 6 r + 15 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 r^{2} + 5 r) + 6 r + 15 = 0$

Factor out the greatest common factor (GCF) from each group.

$r (2 r + 5) + 3 (2 r + 5) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 r + 5$ .

$(2 r + 5) (r + 3) = 0$

Step 3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 r + 5 = 0$

$r + 3 = 0$

Step 4

Set $2 r + 5$ equal to $0$ and solve for $r$ .

Set $2 r + 5$ equal to $0$ .

$2 r + 5 = 0$

Solve $2 r + 5 = 0$ for $r$ .

Subtract $5$ from both sides of the equation.

$2 r = - 5$

Divide each term in $2 r = - 5$ by $2$ and simplify.

Divide each term in $2 r = - 5$ by $2$ .

$\frac{2 r}{2} = \frac{- 5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 r}{2} = \frac{- 5}{2}$

Divide $r$ by $1$ .

$r = \frac{- 5}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$r = - \frac{5}{2}$

Step 5

Set $r + 3$ equal to $0$ and solve for $r$ .

Set $r + 3$ equal to $0$ .

$r + 3 = 0$

Subtract $3$ from both sides of the equation.

$r = - 3$

Step 6

The final solution is all the values that make $(2 r + 5) (r + 3) = 0$ true.

$r = - \frac{5}{2}, - 3$

Solve by Factoring 11r+15=-2r^2

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